## Question 2

If P and Q are positive integers, is the product 3P^{Q} divisible by 2?

- 6Q
^{3}+ 2 is an even number - P + 8Q
^{2}is a prime number

### Correct Answer

B

### Solution

**Steps 1 & 2: Understand Question and Draw Inferences**

3P^{Q }is divisible by 2, if:

- 3P
^{Q }is even

–> P^{Q} is even (Odd term 3 plays no role in the even-odd nature of product 3P^{Q})

–> P is even (Power doesn’t impact the even-odd nature of a term)

So, to answer the question we need to find if P is even

**Step 3: Analyze Statement 1**

6Q^{3} + 2 is an even number

Not Sufficient. We do not know if P is even or odd

**Step 4: Analyze Statement 2**

P + 8Q^{2} is a prime number

All the prime numbers except 2 are odd

–> As Q ≠0, P + 8Q^{2}> 2 (Given: Q is a positive integer => Q >0)

–> P + 8Q^{2} is always odd

8Q^{2} is always even

–> P must be odd (Odd + Even = Odd)

Sufficient.

**Step 5: Analyze Both Statements Together (if needed)**

We get a unique answer in step 4, so this step is not required

**Answer: Option (B)**

Hey, how can the answer be B. B gives us information about P whereas 3PQ can be divisible by 2 if Q is even and P is odd. Since we have no information about Q, should the answer not be E?

Okay got it!. The powers were shown as multiplication on my android phone.

Hi , For statement 2, I understand Q is a positive integer i.e. Q >0, however how did we deduce that P + 8Q^2> 2 ??

Hi JIA,

If you look at the question, it said P+8Q”2 is a prime number. All prime numbers are odd except for 2 hence the expression P+8Q”2 > 2 because we know that it cannot be equals to 2. The trick is focusing on the constraints in the question. Once you get that out of the way you can deduce that 8Q”2 will always be even leaving P to definitely be odd.